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To evaluate an Ising partition function in one dimension, we can make use of a clever expansion. First, in general we have

$$e^x = \cosh x + \sinh x$$
$$\cosh(-x) = \cosh x, \sinh(-x) = - \sinh x$$

now if the argument of the exponential is $\sigma \sigma' K$, where $\sigma, \sigma' = \pm 1$ represent allowed spin values and $K$ is a constant, then we have

$$e^{\sigma \sigma' K} = \cosh K ( 1 + \sigma \sigma' \tanh K )$$

hence for a Hamiltonian

$$\mathcal{H} = - k \sum_{\langle ij \rangle} \sigma_i \sigma_j$$

comprising pairwise interactions between nearest neighbour spins, the partition function can be written

$$Z_N = \sum_{\sigma_1 = \pm 1} \sum_{\sigma_2 = \pm 1} \cdots \sum_{\sigma_N = \pm 1} \prod_{\langle ij \rangle} \cosh^N K [ 1 + \sigma_i \sigma_j \tanh K ]$$

Now consider the case of the a ring of $N$ spins (i.e. a 1D Ising magnet with periodic boundaries) the partition function

$$Z_N = \sum_{\sigma_i = \pm 1} \cosh^N K ( 1 + \sigma_1 \sigma_2 \tanh K ) ( 1 + \sigma_2 \sigma_3 \tanh K ) \cdots ( 1 + \sigma_N \sigma_1 \tanh K )$$

the summand can be organised in powers of $\tanh K$, with coefficients of products and sums of $\sigma$s. Now because $\sigma_i = \pm 1$, products reduce according to $(\sigma_i)^2 = 1$, hence when summed over those terms containing an odd number of $\sigma_i$s will vanish, e.g.

$$\sum_{\sigma_1 = \pm 1} \sigma_1 \sigma_2 = \sigma_2 - \sigma_2 = 0$$

so in general, for a ring

$$Z_N = 2^N [ \cosh^N K + \sinh^N K ]$$

where the factor of $2^N$ comes from $N$ nested sums.

To compute the $Z_a$, notice that the network consists of two rings with $N=3$, joined at one site $\sigma_3$. Each triangle will thus have a partition function

$$Z_3(\sigma_3) = \sum_{\sigma_1 = \pm 1} \sum_{\sigma_3 = \pm 1} \cosh^N K ( 1 + \sigma_1 \sigma_2 \tanh K ) ( 1 + \sigma_2 \sigma_3 \tanh K ) \cdots ( 1 + \sigma_N \sigma_1 \tanh K )$$

i.e. the same as $Z_3$, but with a possible dependence upon $\sigma_2$, since it is not summed over. The total partition function $Z_a$ then factorises into

$$Z_a = \sum_{\sigma_2 = \pm 1} Z_3(\sigma_2) Z_3(\sigma_2)$$

If we expand and sum over $\sigma_1, \sigma_3$ in $Z_3(\sigma_2)$, we notice that the $\sigma_2$ dependence actually cancels, hence $Z'_3 := Z_3(\sigma_2) = Z_3 / 2$ where the factor of $1/2$ accounts for the absence of the $\sigma_2$ sum. The partition function is thus

$$Z_a = 2 {Z'}^2_3 = Z^2_3 / 2 = 2^5 [ \cosh^3 K + \sinh^3 K ]^2$$

The partition function for b is simply a ring with $N=6$, i.e.

$$Z_b = 2^6 [ \cosh^6 K + \sinh^6 K ]$$

We can use a similar approach with c, noting the factorisation

$$Z_c = \sum_{\sigma_2 = \pm 1} \sum_{\sigma_5 = \pm 1} Z_2(\sigma_2) Z_2(\sigma_5) Z_4(\sigma_2, \sigma_5)$$

where

$$Z_2(\sigma_2) = \sum_{\sigma_3 = \pm 1} e^{-K \sigma_3 \sigma_2} = 2 \cosh(K \sigma_2) = 2 \cosh K$$

independent of $\sigma_2$, since $\cosh$ is an even function. We obtain a similar result for $Z_2(\sigma_5)$. Evaluating $Z_4(\sigma_2,\sigma_5)$ is more complicated. We note that the expansion will involve terms up to $\tanh^4 K$, whch by the same argument as $N=3$ will have a coefficient of unity. Now in general only $\sigma_2 \sigma_5$ terms will survive the summation, for example in the linear term

$$(\sigma_1 \sigma_2 + \sigma_2 \sigma_5 + \sigma_5 \sigma_6 + \sigma_6 \sigma_1) \tanh K \rightarrow \sigma_2 \sigma_5 \tanh K$$

The squared term, after eliminating $(\sigma_i)^2 = 1$ terms, becomes

$$(\sigma_1 \sigma_5 + \sigma_2 \sigma_6 + \sigma_5 \sigma_1 + \sigma_6 \sigma_2 + \sigma_1 \sigma_2 \sigma_5 \sigma_6 + \sigma_2 \sigma_5 \sigma_6 \sigma_1) \tanh^2 K \rightarrow 0$$

and the cubed term, after eliminations, will have the same coefficients as the linear term, hence only the term

$$\sigma_2 \sigma_5 \tanh^3 K$$

will remain. We are left with

$$Z_4(\sigma_2, \sigma_5) = 2^2 \cosh^4 K [ 1 + (\tanh K + \tanh^3 K) \sigma_2 \sigma_5 + \tanh^4 K ]$$

thus the overall partition function is

$$Z_c = \sum_{\sigma_2 = \pm 1} \sum_{\sigma_5 = \pm 1} 2^2 \cosh^2 K 2^2 \cosh^4 K [ 1 + \sigma_2 \sigma_5 (\tanh K + \tanh^3 K) + \tanh^4 K ]$$
$$= 2^6 \cosh^6 K [ 1 + \tanh^4 K ]$$

where under summation we keep only those terms that do not involve $\sigma$s. So, to summarise

$$Z_a = 2^5 [ \cosh^3 K + \sinh^3 K ]^2$$
$$Z_b = 2^6 [ \cosh^6 K + \sinh^6 K ]$$
$$Z_c = 2^6 \cosh^6 K [ 1 + \tanh^4 K ]$$

in general we note that there is one power of $2$ for each spin site (graph node), and one power of $\cosh K$ for each interaction (graph vertex) in each partition function.

Now to compare the free energies

$$\mathcal{F} = - k_B T \log Z$$

we can take ratios of the partition functions (since $\log$ is monotonic function).

$$\frac{Z_a}{Z_b} = \frac{1}{2} \left[ 1 + 2 \frac{\cosh^3 K \sinh^3 K}{ \cosh^6 K + \sinh^6 K } \right] = \frac{1}{2} + \frac{x^3}{ 1 + x^6 } < 1 \Rightarrow Z_a > Z_b$$

where we have defined $x = \tanh K < 1$, and used the fact that

$$\frac{x}{1+x^2} < \frac{1}{2} \, ; \, 0 \leq x < 1$$

Now comparing $Z_b$ and $Z_c$

$$\frac{Z_b}{Z_c} = \frac{1 + x^6}{1 + x^4} < 1 \Rightarrow Z_b < Z_c$$

so we still need to compare $Z_a$ and $Z_c$

$$\frac{Z_a}{Z_c} = \frac{1}{2} \frac{ (1 + x^3)^2 }{ 1 + x^4 } < 1 \Rightarrow Z_c > Z_a$$

as can be seen by considering

$$\left. \frac{Z_a}{Z_c} \right|_{x=0} = \frac{1}{2}, \left. \frac{Z_a}{Z_c} \right|_{x=1} = 1$$

and the function has no zeroes or singularities in $[0,1]$. Hence we have

$$Z_c > Z_a > Z_b$$

and since the free energy is negative

$$\mathcal{F}_b > \mathcal{F}_a > \mathcal{F}_c$$

The 2-point correlation function is defined by

$$\langle \sigma_1 \sigma_2 \rangle = \frac{1}{Z_N} \sum_\mathrm{conf} \sigma_1 \sigma_3 e^{-\beta \mathcal{H}}$$

for (a) we make use of the same factorisation used to compute the partition function

$$\langle \sigma_1 \sigma_2 \rangle_a = \frac{1}{Z_a} \sum_{\sigma_2 = \pm 1} \left[ \sum_{\sigma_1 = \pm 1} \sigma_1 \sum_{\sigma_5 = \pm 1} \exp \{ K( \sigma_1 \sigma_2 + \sigma_2 \sigma_5 + \sigma_5 \sigma_1 ) \} \right]^2$$

examining the expansion of the inner sum over $\sigma_5$

\begin{eqnarray*}
S(\sigma_1, \sigma_2) &=& \sum_{\sigma_5 = \pm 1} \cosh^3 K ( 1 + (\tanh K + \tanh^2 K) ( \sigma_1 \sigma_2 + \sigma_2 \sigma_5 + \sigma_5 \sigma_1 ) + \tanh^3 K ) \\
&=& 2 \cosh^3 K ( 1 + (\tanh K + \tanh^2 K) \sigma_1 \sigma_2 + \tanh^3 K )
\end{eqnarray*}

the next sum, over $\sigma_1$, yields

$$S(\sigma_2) = \sum_{\sigma_1 = \pm 1} \sigma_1 S(\sigma_1, \sigma_2) = 2^2 \cosh^3 K (\tanh K + \tanh^2 K) \sigma_2$$

and so the full correlation function is

$$\langle \sigma_1 \sigma_3 \rangle_a = \frac{1}{Z_a} \sum_{\sigma_2 = \pm 1} S(\sigma_2)^2 = \frac{1}{Z_a} 2^5 \cosh^6 K (\tanh K + \tanh^2 K)^2$$

$$\Rightarrow \langle \sigma_1 \sigma_3 \rangle_a = x^2 \frac{(1 + x)^2 }{ (1+x^3)^2 } \, ; \, x = \tanh K$$

To evaluate (b) we can use the transfer matrix approach detailed in the notes

$$\langle \sigma_1 \sigma_{n+1} \rangle = \frac{ \lambda_1^n \lambda_2^{N-n} + \lambda_2^n \lambda_1^{N-n} }{ \lambda_1^N + \lambda_2^N }$$

where $N=6$, $n=2$ and $\lambda_1 = 2 \cosh K$, $\lambda_2 = 2 \sinh K$, giving

$$\langle \sigma_1 \sigma_3 \rangle_b = \cosh^2 K \sinh^2 K \frac{ \sinh^2 K + \cosh^2 K }{ \sinh^6 K + \cosh^6 K } = x^2 \frac{ 1 + x^2 }{ 1 + x^6 }$$

Finally for (c) we use a similar factorisation approach

$$\langle \sigma_1 \sigma_3 \rangle_c = \frac{1}{Z_c} \sum_{\sigma_2 = \pm 1} S_2(\sigma_2) \sum_{\sigma_5 = \pm 1} S_4(\sigma_2,\sigma_5) S_2(\sigma_5)$$

The ``leg'' factors are given by

$$S_2(\sigma_5) = 2 \cosh K$$

$$S_2(\sigma_2) = \sum_{\sigma_3 = \pm 1} \sigma_3 e^{K\sigma_3 \sigma_2} = 2 \sinh K \sigma_2 = 2 \sigma_2 \sinh K$$

the square $S_4$ factor is rather more complicated

\begin{eqnarray*}
S_4(\sigma_2,\sigma_5) &=& \sum_{\sigma_1 = \pm 1} \sum_{\sigma_6 = \pm 1} \sigma_1 \exp \{ K( \sigma_2 \sigma_5 + \sigma_5 \sigma_6 + \sigma_6 \sigma_1 + \sigma_1 \sigma_2 ) \} \\
&=& \sum_{\sigma_1 = \pm 1} \sigma_1 e^{ K( \sigma_2 \sigma_5 + \sigma_1 \sigma_2 ) } \sum_{\sigma_6 = \pm 1} e^{ K( \sigma_5 \sigma_6 + \sigma_6 \sigma_1 ) } \\
&=& \sum_{\sigma_1 = \pm 1} \sigma_1 e^{ K( \sigma_2 \sigma_5 + \sigma_1 \sigma_2 ) } \sum_{\sigma_6 = \pm 1} \cosh^2 K ( 1 + \tanh K ( \sigma_5 \sigma_6 + \sigma_6 \sigma_1 ) + \sigma_5 \sigma_1 \tanh^2 K ) \\
&=& 2 \cosh^2 K \sum_{\sigma_1 = \pm 1} \sigma_1 e^{ K( \sigma_2 \sigma_5 + \sigma_1 \sigma_2 ) } ( 1 + \sigma_5 \sigma_1 \tanh^2 K ) \\
&=& 2 \cosh^2 K e^{ K \sigma_2 \sigma_5 }\sum_{\sigma_1 = \pm 1} \sigma_1 \cosh K ( 1 + \sigma_1 \sigma_2 \tanh K ) ( 1 + \sigma_5 \sigma_1 \tanh^2 K ) \\
&=& 2 \cosh^3 K e^{ K \sigma_2 \sigma_5 }\sum_{\sigma_1 = \pm 1} \sigma_1 ( \sigma_1 + \sigma_2 \tanh K + \sigma_5 \tanh^2 K + \sigma_1 \sigma_2 \sigma_5 \tanh^3 K ) \\
&=& 2^2 \cosh^4 K ( 1 + \sigma_2 \sigma_5 \tanh K ) (\sigma_2 \tanh K + \sigma_5 \tanh^2 K ) \\
\end{eqnarray*}

$$\langle \sigma_1 \sigma_3 \rangle_c = \frac{1}{Z_c} 2^6 \sinh K \cosh^5 K (\tanh K + \tanh^3 K)$$

$$\Rightarrow \langle \sigma_1 \sigma_3 \rangle_c = \sinh K \cosh^5 K \frac{ \tanh K + \tanh^3 K }{ \cosh^6 (1 + \tanh^4 K) } = x^2 \frac{1+x^2}{1+x^4}$$

Again, we can use ratios to compare correlation functions

$$\frac{ \langle \sigma_1 \sigma_3 \rangle_a }{ \langle \sigma_1 \sigma_3 \rangle_b } = \frac{(1 + x)^2}{(1+x^3)^2} \frac{1+x^2}{1+x^6} > 1 \Rightarrow \langle \sigma_1 \sigma_3 \rangle_a > \langle \sigma_1 \sigma_3 \rangle_b$$

$$\frac{ \langle \sigma_1 \sigma_3 \rangle_b }{ \langle \sigma_1 \sigma_3 \rangle_c } = \frac{1 + x^4}{1+ x^6} > 1 \Rightarrow \langle \sigma_1 \sigma_3 \rangle_b > \langle \sigma_1 \sigma_3 \rangle_c$$

So for finite temperature $0 < x < 1$

$$\langle \sigma_1 \sigma_3 \rangle_a > \langle \sigma_1 \sigma_3 \rangle_b > \langle \sigma_1 \sigma_3 \rangle_c$$

We can understand this intuitively in terms of the number of ``routes'' connecting the nodes. In (a) there are a 4 acyclic routes connecting $\sigma_1$ with $\sigma_3$, all via $\sigma_2$; 1 with 2 ``hops'', 2 with 3 hops, and 1 with 4 hops. In (b) there are 2 routes, 1 with 2 hops and 1 with 4. In (c) there is also 1 2-hop route and 1 4-hop route, but the ``influence'' of these routes is decreased by the increased coupling to other nodes. 

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